Difference between revisions of "2010 AMC 10A Problems/Problem 11"
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+ | == Problem 11 == | ||
+ | The length of the interval of solutions of the inequality <math>a \le 2x + 3 \le b</math> is <math>10</math>. What is <math>b - a</math>? | ||
+ | |||
+ | <math> | ||
+ | \mathrm{(A)}\ 6 | ||
+ | \qquad | ||
+ | \mathrm{(B)}\ 10 | ||
+ | \qquad | ||
+ | \mathrm{(C)}\ 15 | ||
+ | \qquad | ||
+ | \mathrm{(D)}\ 20 | ||
+ | \qquad | ||
+ | \mathrm{(E)}\ 30 | ||
+ | </math> | ||
+ | |||
+ | |||
+ | ==Solution== | ||
+ | |||
Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. | Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. | ||
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We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math> | We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math> | ||
+ | |||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}} |
Revision as of 09:54, 2 January 2012
Problem 11
The length of the interval of solutions of the inequality is . What is ?
Solution
Since we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .
Subtract from all of the quantities:
Divide all of the quantities by .
Since we have the range of the solutions, we can make them equal to .
Multiply both sides by 2.
Re-write without using parentheses.
Simplify.
We need to find for the problem, so the answer is
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |